Question: In right triangle $DEF$, we have $\sin D = \frac{5}{13}$ and $\sin E = 1$.  Find $\sin F$.
Explanation: Since $\sin E = 1$, we have $\angle E = 90^\circ$, so our triangle is as shown below:

[asy]
pair D,EE,F;

EE = (0,0);
F = (5,0);
D = (0,12);
draw(D--EE--F--D);
draw(rightanglemark(F,EE,D,18));
label("$E$",EE,SW);
label("$F$",F,SE);
label("$D$",D,N);
[/asy]

Since $\sin D = \frac{5}{13}$, we have $\frac{EF}{DF} = \frac{5}{13}$, so $\cos F = \frac{EF}{DF} = \frac{5}{13}$.  Since $\sin^2 F + \cos^2 F = 1$, and $\angle F$ is acute (so $\sin F$ is positive), we have \[\sin F =\sqrt{1 - \cos^2 F} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \boxed{\frac{12}{13}}.\]We also could have noticed that since $\frac{EF}{DF} = \frac{5}{13}$, we have $EF = 5x$ and $DF = 13x$ for some value of $x$.  Then, from the $\{5,12,13\}$ Pythagorean triple, we see that $DE = 12x$, so $\sin F = \frac{DE}{DF} = \frac{12}{13}$.